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I have put his on wolfram and obtained answer as follows:

$\sum_{n=1}^{\infty}\frac{n}{2^n} = 2$

And the series is convergent too because $\lim_{n\to\infty} \frac {n}{2^n} = 0$

However I am wondering if there is a convenient way to solve this; I don't think you can represent it by a geometric progression either. So how do we have to do it on paper?

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    $\begingroup$ Of course the fact that $\lim_{n \to \infty} \frac{n}{2^n} \to 0$ doesn't imply that the series is convergent. $\endgroup$
    – Amateur
    Apr 17, 2014 at 2:49
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    $\begingroup$ Differentiate the geometric series. $\endgroup$
    – MattyZ
    Apr 17, 2014 at 2:50

5 Answers 5

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Let $S$ be our sum. Then $S=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\cdots$. Thus $$\begin{align}2S&=1+&\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}+\cdots\\ S&=&\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\cdots\end{align} $$ Subtract. We get $$S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\cdots=2.$$

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Hint:

$$ \sum_{k=1}^{\infty} x^k = \frac{x}{1-x} \implies \sum_{k=1}^{\infty} kx^k=\frac{d}{dx} \frac{x}{1-x} . $$

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  • $\begingroup$ Want to understand why we can differentiate. $\endgroup$
    – deostroll
    Apr 17, 2014 at 4:14
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    $\begingroup$ @deostroll: We can differentiate a power series within its radius of convergence. $\endgroup$ Apr 17, 2014 at 4:21
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    $\begingroup$ @deostroll The technical result allowing us to do this is the following: First, power series $\sum \alpha_n x^n$ converge absolutely and uniformly on any closed subset of their (open) interval of convergence. (This follows from the $M$-test.) Second, the (formal) series of their termwise derivative, $\sum n\alpha_n x^{n-1}$, has the same radius of convergence. (Cont.) $\endgroup$ Apr 19, 2014 at 3:11
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    $\begingroup$ @deostroll Third, and this is the technical part, if a sequence $f_n$ of differentiable functions defined on some closed interval satisfies that (1) $f_n'\to g$ uniformly, and (2) $f_n(x_0)$ converges (as $n\to\infty)$ for some $x_0$, then (a) $f_n$ converges uniformly to some differentiable function $f$, and (b) $f'=g$. To apply this, take as $f_n$ the partial sums of the given power series. A proof of this theorem appears for instance in Rudin's Principles of mathematical analysis, see Theorem 7.17. (Cont.) $\endgroup$ Apr 19, 2014 at 3:15
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    $\begingroup$ @deostroll The assumption is technical, but cannot really be relaxed: Even if we assume that $f_n\to f$ uniformly, that $f$ is differentiable, and that $f_n'\to g$ pointwise, we may have that $f'(x)\ne g(x)$ for all $x$. This is a (remarkably recent!) theorem of Darji, see here. $\endgroup$ Apr 19, 2014 at 3:17
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Another way to do the problem is to note that $$\left(\sum_{k=1}^\infty \frac{1}{2^k}\right)^2 = \sum_{k=1}^\infty \frac{k-1}{2^k}$$ so that the desired series is really just $$\left(\sum_{k=1}^\infty \frac{1}{2^k}\right)^2 + \sum_{k=1}^\infty \frac{1}{2^k}$$

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If $\displaystyle S=\sum_{n=1}^\infty\frac{n}{2^n}$, then $$S-\frac12=\sum_{n=1}^\infty\frac{n+1}{2^{n+1}}=\frac12\sum_{n=1}^\infty\frac{n+1}{2^n}=\frac12\left(S+\sum_{n=1}^\infty\frac1{2^n}\right)=\frac12(S+1),$$ so $2S-1=S+1$, or $S=2$.

All that remains is to justify that the series converges. But $n<1.1^n$ for $n$ large enough, so a tail of $S$ is bounded above by a tail of the convergent geometric series $\sum_n\left(\frac{1.1}2\right)^n$.

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In this answer we go for the 'one trick pony' approach - all we know is that for $k \ge 0$,

$\tag 1 \sum_{n=k}^{\infty}\frac{1}{2^n} = 2^{1-k}$

We decompose the summands of $\sum_{n=1}^{\infty}\frac{n}{2^n}$ in a natural/straightforward manner, arranging these numbers into a table:

$$\begin{pmatrix} \frac{1}{2} & \frac{1}{4} & \frac{1}{8} & \frac{1}{16} & \frac{1}{32} & \dots \\ 0 & \frac{1}{4} & \frac{1}{8} & \frac{1}{16} & \frac{1}{32} & \dots \\ 0 & 0 & \frac{1}{8} & \frac{1}{16} & \frac{1}{32} & \dots \\ 0 & 0 & 0 & \frac{1}{16} & \frac{1}{32} & \dots \\ 0 & 0 & 0 & 0 & \frac{1}{32} & \dots \\ 0 & 0 & 0 & 0 & 0 & \dots \\ . \\ . \\ . \\ \end{pmatrix}$$

Using $\text{(1)}$ we add up the entries in each row,

$$\begin{pmatrix} 1 \\ \frac{1}{2} \\ \frac{1}{4} \\ \frac{1}{8} \\ \frac{1}{16} \\ . \\ . \\ . \\ \end{pmatrix}$$

And now we add up the entries of our column vector, giving

$\quad \text{ANS: } 2$

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  • $\begingroup$ Hey, it was good enough for Jacob Bernoulli... [see, for instance, Section 4.1 of digitalcommons.ursinus.edu/cgi/… ] $\endgroup$
    – user882145
    May 13, 2023 at 5:03

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